151116 初版 151116 更新

\(\tan\left(2\theta+\dfrac{\pi}{3}\right)=1\) を満たす θ
一般には,
\(2\theta+\dfrac{\pi}{3} = \dfrac{\pi}{4}+n\pi\)   (n は整数)
すなわち,
\(\theta = \dfrac{-1+12n}{24}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\theta = \dfrac{11}{24}\pi\),  \(\dfrac{23}{24}\pi\),   \(\dfrac{35}{24}\pi\),   \(\dfrac{47}{24}\pi\)  

\(\tan\left(2\theta+\dfrac{\pi}{3}\right) > 1\) を満たす θ
一般には,
\(\dfrac{\pi}{4}+n\pi < 2\theta+\dfrac{\pi}{3}\) \(< \dfrac{\pi}{2}+n\pi\)   (n は整数)
すなわち,
\(\dfrac{-1+12n}{24}\pi < \theta\) \(< \dfrac{1+6n}{12}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(0\leqq \theta < \dfrac{\pi}{12}\),   \(\dfrac{11}{24}\pi < \theta < \dfrac{7}{12}\pi\),   \(\dfrac{23}{24}\pi < \theta < \dfrac{13}{12}\pi\),  
\(\dfrac{35}{24}\pi < \theta < \dfrac{19}{12}\pi\),   \(\dfrac{47}{24}\pi < \theta < 2\pi\)  

\(\tan\left(2\theta+\dfrac{\pi}{3}\right) < 1\) を満たす θ
一般には,
\(-\dfrac{\pi}{2}+n\pi < 2\theta+\dfrac{\pi}{3}\) \(< \dfrac{\pi}{4}+n\pi\)   (n は整数)
すなわち,
\(\dfrac{-5+6n}{12}\pi < \theta\) \(< \dfrac{-1+12n}{24}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\dfrac{\pi}{12} < \theta < \dfrac{11}{24}\pi\),   \(\dfrac{7}{12}\pi < \theta < \dfrac{23}{24}\pi\),  
\(\dfrac{13}{12}\pi < \theta < \dfrac{35}{24}\pi\),   \(\dfrac{19}{12}\pi < \theta < \dfrac{47}{24}\pi\)  

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